A) 75
B) 60
C) 80
D) 70
Correct Answer: B
Solution :
\[i=\frac{Normal\text{ }molar\text{ }mass}{Abnormal\text{ }molar\text{ }mass}\]=\[\frac{\left( 40\,g\,mo{{l}^{-1}} \right)}{\left( 25\,g\,mo{{l}^{-1}} \right)}\]=1.6 \[\alpha =\frac{i-1}{n-1}=\frac{1.6-1}{2-1}=0.6\] % ionisation =60.You need to login to perform this action.
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